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Let $G$ be a group of order $399$. Then by Sylow, it must have a normal subgroup of order $19$, denoted by $H$. Let $N, K$ be groups of order $7$ and $3$. Then $HN$ and $HK$ are groups of order $133$ and $57$. What is the possible order of their normalizer?

The normalizer can only be either themselves or the whole group by Lagrang

zach
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Without knowing more about $G$, both cases are possible for $HK$, but $HN$ is normal in $G$. Since $HN$ is of index 3 in $G$, it contains a normal subgroup (the normal core) of index $\leq 3!=6$ so must be $HN$ itself.

Consider the group $G=F_{21}\times C_{19}$ where $F_{21}$ is the nonabelian group of order 21. A subgroup $C_3\times C_{19}$ cannot be normal in $G$, or else the $C_3$ would be a characteristic subgroup of a normal subgroup, hence normal subgroup of $G$, contradicting $F_{21}$ nonabelian. So $N_G(HK)=HK$ in this case.

On the other hand, the abelian group of order 399 has $N_G(HK)=G$.

user10354138
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  • what is a normal core? – zach Jun 16 '19 at 16:26
  • Normal core of a subgroup – user10354138 Jun 16 '19 at 16:31
  • Is there a more elementary approach? This seems too advanced for me. I am thinking: $G = F_{57} \times Z_7$ if the subgroup of order $57$ is normal, but do not know how to derive a contradiction. – zach Jun 16 '19 at 22:15
  • Subgroups of order 57 need not be normal, as the $F_{21}\times C_{19}$ example shows. Indeed there are only two groups of order 399, the abelian one and this, because we can prove the Sylow-7 is normal (there is only one Sylow-7 in the normal subgroup of order 133, so characteristic in normal, hence normal). – user10354138 Jun 17 '19 at 05:36